On a strengthened Hardy-Hilbert type inequality

Abstract

We derive a strengthenment of a Hardy-Hilbert type inequality by using the Euler-Maclaurin expansion for the zeta function and estimating the weight function effectively. As applications, some particular results are presented.

MSC:26D15.

1 Introduction

Let $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $a_n,b_n\ge 0$, $0<{\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{b}_n^q<\mathrm{\infty }$. Then one [1] has

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{m+n}<\frac{\pi }{sin(\pi /p)}{\left[\sum _{n=1}^{\mathrm{\infty }}{a}_n^p\right]}^{\frac{1}{p}}{\left[\sum _{n=1}^{\mathrm{\infty }}{b}_n^q\right]}^{\frac{1}{q}},$$

(1.1)

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m,n\}}<pq{\left[\sum _{n=1}^{\mathrm{\infty }}{a}_n^p\right]}^{\frac{1}{p}}{\left[\sum _{n=1}^{\mathrm{\infty }}{b}_n^q\right]}^{\frac{1}{q}},$$

(1.2)

where the constant factor $\frac{\pi }{sin(\pi /p)}$ and pq are best possible. Inequality (1.1) is well known as Hardy-Hilbert’s inequality, and inequality (1.2) is named a Hardy-Hilbert type inequality. Both of them are important in analysis and applications [2]. In recent years, many results about generalizations of this type of inequality were established (see [3]). Under the same conditions as (1.1) and (1.2), some Hardy-Hilbert type inequalities, which are similar to (1.1) and (1.2), have been studied and generalized by some mathematicians.

By introducing a parameter, Yang gave a generalization of inequality (1.2) with the best constant factor as follows:

If $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $2-min\{p,q\}<\lambda \le 2$, $a_n,b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{b}_n^q<\mathrm{\infty }$, then

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<k_{\lambda }(p){\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{b}_n^q\right\}}^{\frac{1}{q}},$$

(1.3)

where the constant factor $k_{\lambda }(p)=\frac{\lambda pq}{(p+\lambda -2)(q+\lambda -2)}$ is best possible.

Furthermore, by introducing a parameter and two pairs of conjugate exponents, Zhong gave a generalization of inequality (1.3) with the best constant factor as follows:

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $0<\lambda \le min\{r,s\}$, $a_n,b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{p(1-\frac{\lambda }{r})-1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{s})-1}{b}_n^q<\mathrm{\infty }$, then

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<k_{\lambda }(r){\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{p(1-\frac{\lambda }{r})-1}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{s})-1}{b}_n^q\right\}}^{\frac{1}{q}},$$

(1.4)

where the constant factor $k_{\lambda }(r)=\frac{rs}{\lambda }$ is best possible.

Recently, in [4], Jiang and Hua established an improvement of inequality (1.3) as follows:

If $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $2-min\{p,q\}<\lambda \le 2$, $a_n\ge 0$, $b_n\ge 0$, for $n\ge 1$, $n\in \mathrm{N}$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{a}_n^p<\mathrm{\infty }$, $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{1-\lambda }{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k(\lambda )-\frac{q}{3(q+\lambda -2)n^{\frac{q+\lambda -2}{q}}}]n^{1-\lambda }{a}_n^p\right\}}^{\frac{1}{p}}\\ \times \left\{\sum _{n=1}^{\mathrm{\infty }}[k(\lambda )-\frac{p}{3(p+\lambda -2)n^{\frac{p+\lambda -2}{p}}}]n^{1-\lambda }{b}_n^q\right\},\end{array}$$

(1.5)

where $k(\lambda )=\frac{pq\lambda }{(p+\lambda -2)(q+\lambda -2)}>0$.

In this paper, by introducing a parameter and estimating the weight coefficient, we obtain a strengthenment of inequality (1.4) and generalize inequality (1.5). As applications, some particular results are presented.

2 Some preliminary results

First, we need the following formula of the Riemann-ζ function (see [5]):

$$\begin{array}{rl}\zeta (\rho )=& \sum _{n=1}^m\frac{1}{n^{\rho }}-\frac{m^{1-\rho }}{1-\rho }-\frac{1}{2m^{\rho }}\\ -\sum _{n=1}^{l-1}\frac{B_{2n}}{2n}\left(\begin{array}{c}-\rho \\ 2n-1\end{array}\right)\frac{1}{m^{\rho +2n-1}}-\frac{B_{2l}}{2l}\left(\begin{array}{c}-\rho \\ 2l-1\end{array}\right)\frac{\epsilon }{m^{\rho +2l-1}},\end{array}$$

(2.1)

where $\rho >0$, $\rho \ne 1$, $m,l\ge 1$, $m,l\in \mathrm{N}$, $0<\epsilon =\epsilon (\rho ,l,m)<1$. The numbers $B_1=-1/2$, $B_2=1/6$, $B_3=0$, $B_4=-1/30$, … are Bernoulli numbers. In particular, $\zeta (\rho )={\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^{\rho }}$ ($\rho >1$).

Since $\zeta (0)=-1/2$, the formula of the Riemann-ζ function (2.1) also holds for $\rho =0$.

Lemma 2.1 Let $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $0<\lambda \le min\{r,s\}$, define the weight coefficients $\omega (m,\lambda ,s)$ and $\omega (n,\lambda ,r)$ as

$$\omega (m,\lambda ,s)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}},$$

(2.2)

$$\omega (n,\lambda ,r)=\sum _{m=1}^{\mathrm{\infty }}\frac{1}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{n}{m}\right)}^{1-\frac{\lambda }{r}}.$$

(2.3)

Then we have

$$\omega (m,\lambda ,s)<m^{1-\lambda }[k_{\lambda }-\frac{s}{3\lambda m^{\frac{\lambda }{s}}}]$$

(2.4)

and

$$\omega (n,\lambda ,r)<n^{1-\lambda }[k_{\lambda }-\frac{r}{3\lambda m^{\frac{\lambda }{r}}}],$$

(2.5)

where $k_{\lambda }=\frac{rs}{\lambda }$.

Proof For $0<\lambda \le min\{r,s\}$, taking $\rho =1-\frac{\lambda }{s}\ge 0$, $l=1$ in (2.1), we get

$$\zeta (1-\frac{\lambda }{s})=\sum _{n=1}^m\frac{1}{n^{1-\frac{\lambda }{s}}}-\frac{s{m}^{\frac{\lambda }{s}}}{\lambda }-\frac{1}{2m^{1-\frac{\lambda }{s}}}+\frac{1-\frac{\lambda }{s}}{12m^{2-\frac{\lambda }{s}}}{\epsilon }_1,$$

(2.6)

where $0<{\epsilon }_1<1$.

Set $\rho =1+\frac{\lambda }{r}$, $l=1$, and we can derive

$$\zeta (1+\frac{\lambda }{r})=\sum _{n=1}^{m-1}\frac{1}{n^{1+\frac{\lambda }{r}}}+\frac{r{m}^{-\frac{\lambda }{r}}}{\lambda }+\frac{1}{2m^{1+\frac{\lambda }{r}}}+\frac{1+\frac{\lambda }{r}}{12m^{2+\frac{\lambda }{r}}}{\epsilon }_2,$$

(2.7)

where $0<{\epsilon }_2<1$.

Thus we get

$$\begin{array}{rl}\omega (m,\lambda ,s)& =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}\\ =\sum _{n=1}^m\frac{1}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}-\frac{1}{m^{\lambda }}+\sum _{n=m}^{\mathrm{\infty }}\frac{1}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}\\ =\sum _{n=1}^m\frac{1}{m^{\lambda }}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}-\frac{1}{m^{\lambda }}+\sum _{n=m}^{\mathrm{\infty }}\frac{1}{n^{\lambda }}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}\\ =\frac{1}{m^{\lambda +\frac{\lambda }{s}-1}}\sum _{n=1}^m\frac{1}{n^{1-\frac{\lambda }{s}}}-\frac{1}{m^{\lambda }}+m^{1-\frac{\lambda }{s}}\sum _{n=m}^{\mathrm{\infty }}\frac{1}{n^{1+\frac{\lambda }{r}}}.\end{array}$$

Combining (2.6) and (2.7), we have

$$\begin{array}{rl}\omega (m,\lambda ,s)<& \frac{1}{m^{\lambda +\frac{\lambda }{s}-1}}[\zeta (1-\frac{\lambda }{s})+\frac{s{m}^{\frac{\lambda }{s}}}{\lambda }+\frac{1}{2m^{1-\frac{\lambda }{s}}}]-\frac{1}{m^{\lambda }}\\ +m^{1-\frac{\lambda }{s}}[\frac{r{m}^{-\frac{\lambda }{r}}}{\lambda }+\frac{1}{2m^{1+\frac{\lambda }{r}}}+\frac{1+\frac{\lambda }{r}}{12m^{2+\frac{\lambda }{r}}}]\\ =& \frac{1}{m^{\lambda +\frac{\lambda }{s}-1}}\zeta (1-\frac{\lambda }{s})+\frac{s{m}^{1-\lambda }}{\lambda }+\frac{1}{2m^{\lambda }}-\frac{1}{m^{\lambda }}+\frac{r{m}^{1-\lambda }}{\lambda }+\frac{1}{2m^{\lambda }}+\frac{1+\frac{\lambda }{r}}{12m^{1+\lambda }}\\ =& \frac{1}{m^{\lambda +\frac{\lambda }{s}-1}}\zeta (1-\frac{\lambda }{s})+\frac{rs{m}^{1-\lambda }}{\lambda }+\frac{1+\frac{\lambda }{r}}{12m^{1+\lambda }}\\ =& m^{1-\lambda }\{\frac{rs}{\lambda }-\frac{1}{m^{\frac{\lambda }{s}}}[-\zeta (1-\frac{\lambda }{s})-\frac{1+\frac{\lambda }{r}}{12m^{2-\frac{\lambda }{s}}}]\}.\end{array}$$

In (2.6), let $m=1$, by $0<\lambda \le min\{r,s\}$, we obtain

$$\begin{array}{rl}\zeta (1-\frac{\lambda }{s})& =1-\frac{s}{\lambda }-\frac{1}{2}+\frac{(1-\frac{\lambda }{s}){\epsilon }_1}{12}<\frac{1}{2}-\frac{s}{\lambda }+\frac{1-\frac{\lambda }{s}}{12}=\frac{6\lambda -12s-\lambda (1-\frac{\lambda }{s})}{12\lambda }\\ <\frac{6\lambda -12s-(\lambda -s)}{12\lambda }=\frac{5\lambda -11s}{12\lambda }=-\frac{11s-5\lambda }{12\lambda }<0.\end{array}$$

Therefore, for $m\ge 1$, $m\in \mathrm{N}$, $0<\lambda \le min\{r,s\}$, we obtain

$$\begin{array}{rl}-\zeta (1-\frac{\lambda }{s})-\frac{1+\frac{\lambda }{r}}{12m^{2-\frac{\lambda }{s}}}& >\frac{11s-5\lambda }{12\lambda }-\frac{1+\frac{\lambda }{r}}{12}=\frac{11s-5\lambda -\lambda (1+\frac{\lambda }{r})}{12\lambda }\ge \frac{11s-5\lambda -2\lambda }{12\lambda }\\ =\frac{4s+7(s-\lambda )}{12\lambda }\ge \frac{4s}{12\lambda }=\frac{s}{3\lambda }.\end{array}$$

Applying the above inequality, we obtain (2.4). Similarly, we can prove (2.5). The lemma is proved. □

3 Main results

Theorem 3.1 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $0<\lambda \le min\{r,s\}$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{p(1-\frac{\lambda }{r})-1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{s})-1}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{s}{3\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]n^{q(1-\frac{\lambda }{s})-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.1)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\frac{p\lambda }{s}-1}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{s}{3\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p,$$

(3.2)

where $k_{\lambda }=\frac{rs}{\lambda }>0$. Inequality (3.1) is equivalent to (3.2). In particular, we have the following equivalent inequalities:

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& k_{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{s}{k_{\lambda }3\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{s})-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.3)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p\lambda }{s}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<k_{\lambda }^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{s}{3k_{\lambda }\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p.$$

(3.4)

Proof From Hölder inequality (see [6]), we have

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}\\ =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}\frac{n^{(\lambda /s-1)/p}}{m^{(\lambda /r-1)/q}}\frac{m^{(\lambda /r-1)/q}}{n^{(\lambda /s-1)/p}}\\ \le {\left\{\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m^p{m}^{p(1-\lambda /r)+\lambda -2}}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{m}{n}\right)}^{1-\frac{\lambda }{s}}\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{b_n^q{n}^{q(1-\lambda /s)+\lambda -2}}{max\{m^{\lambda },n^{\lambda }\}}{\left(\frac{n}{m}\right)}^{1-\frac{\lambda }{r}}\right\}}^{\frac{1}{q}}\\ ={\{\sum _{m=1}^{\mathrm{\infty }}\omega (m,\lambda ,s)m^{p(1-\lambda /r)+\lambda -2}{a}_m^p\}}^{\frac{1}{p}}{\{\sum _{n=1}^{\mathrm{\infty }}\omega (n,\lambda ,r)n^{q(1-\lambda /s)+\lambda -2}{b}_n^q\}}^{\frac{1}{q}}.\end{array}$$

Hence, by (2.4), (2.5), inequality (3.1) is true.

Setting $b_n$ as

$$b_n=\frac{n^{p\lambda /s-1}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^{p-1},$$

by using (3.1), we have

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]n^{q(1-\frac{\lambda }{s})-1}{b}_n^q\\ =\sum _{n=1}^{\mathrm{\infty }}\frac{n^{p\lambda /s-1}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p\\ =\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}\le {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{s}{3\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]n^{q(1-\frac{\lambda }{s})-1}{b}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

(3.5)

Hence, we obtain

$$\begin{array}{rl}0& <\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\frac{p\lambda }{s}-1}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p\\ <\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{s}{3\lambda n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p<\mathrm{\infty }.\end{array}$$

(3.6)

By (3.1), both (3.5) and (3.6) take the form of strict inequality, and we have (3.2).

On the other hand, suppose that (3.2) is valid, from Hölder inequality, we find

$$\begin{array}{r}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}\\ =\sum _{n=1}^{\mathrm{\infty }}\frac{n^{[q(\lambda /s-1)+1]/q}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{\frac{1}{q}}}\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{\frac{1}{q}}{n}^{[q(1-\lambda /s)-1]/q}{b}_n\\ \le {\left\{\sum _{n=1}^{\mathrm{\infty }}\frac{n^{p\lambda /s-1}}{{[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{r}{3\lambda n^{\frac{\lambda }{r}}}]n^{q(1-\lambda /s)-1}{b}_n^q\right\}}^{\frac{1}{q}}.\end{array}$$

Then, by using (3.2), we have (3.1). Hence, (3.2) and (3.1) are equivalent. The proof of Theorem 3.1 is completed. □

Since $0<\lambda \le min\{r,s\}$, by Theorem 3.1, we have the following.

Corollary 3.2 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $0<\lambda \le min\{r,s\}$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{p(1-\frac{\lambda }{r})-1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{s})-1}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{1}{3n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{1}{3n^{\frac{\lambda }{r}}}]n^{q(1-\frac{\lambda }{s})-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.7)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\frac{p\lambda }{s}-1}}{{[k_{\lambda }-\frac{1}{3n^{\frac{\lambda }{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{1}{3n^{\frac{\lambda }{s}}}]n^{p(1-\frac{\lambda }{r})-1}{a}_n^p,$$

(3.8)

where $k_{\lambda }=\frac{rs}{\lambda }>0$. Inequality (3.7) is equivalent to (3.8).

For $r=s=2$, by using (3.1) and (3.2), we have the following.

Corollary 3.3 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\lambda \le 2$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{p(1-\frac{\lambda }{2})-1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{2})-1}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{2}{3\lambda n^{\frac{\lambda }{2}}}]n^{p(1-\frac{\lambda }{2})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{2}{3\lambda n^{\frac{\lambda }{2}}}]n^{q(1-\frac{\lambda }{2})-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.9)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\frac{p\lambda }{2}-1}}{{[k_{\lambda }-\frac{2}{3\lambda n^{\frac{\lambda }{2}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{2}{3\lambda n^{\frac{\lambda }{2}}}]n^{p(1-\frac{\lambda }{2})-1}{a}_n^p,$$

(3.10)

where $k_{\lambda }=\frac{4}{\lambda }>0$. Inequality (3.9) is equivalent to (3.10). In particular, we have the equivalent inequalities as follows.

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& k_{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{2}{k_{\lambda }3\lambda n^{\frac{\lambda }{2}}}]n^{p(1-\frac{\lambda }{2})-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q(1-\frac{\lambda }{2})-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.11)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p\lambda }{2}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<k_{\lambda }^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{2}{3k_{\lambda }\lambda n^{\frac{\lambda }{2}}}]n^{p(1-\frac{\lambda }{2})-1}{a}_n^p.$$

(3.12)

For $r=q$, $s=p$, by using (3.1) and (3.2), we have the following.

Corollary 3.4 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\lambda \le min\{p,q\}$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{(p-1)(1-\lambda )}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{(q-1)(1-\lambda )}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{p}{3\lambda n^{\frac{\lambda }{p}}}]n^{(p-1)(1-\lambda )}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{q}{3\lambda n^{\frac{\lambda }{q}}}]n^{(q-1)(1-\lambda )}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.13)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\lambda -1}}{{[k_{\lambda }-\frac{q}{3\lambda n^{\frac{\lambda }{q}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{p}{3\lambda n^{\frac{\lambda }{p}}}]n^{(p-1)(1-\lambda )}{a}_n^p,$$

(3.14)

where $k_{\lambda }=\frac{pq}{\lambda }>0$. Inequality (3.13) is equivalent to (3.14). In particular, we have the equivalent inequalities as follows.

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& k_{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{p}{k_{\lambda }3\lambda n^{\frac{\lambda }{p}}}]n^{(p-1)(1-\lambda )}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{(q-1)(1-\lambda )}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.15)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\lambda -1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<k_{\lambda }^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{p}{3k_{\lambda }\lambda n^{\frac{\lambda }{p}}}]n^{(p-1)(1-\lambda )}{a}_n^p.$$

(3.16)

For $r=p$, $s=q$, by using (3.1) and (3.2), we have the following.

Corollary 3.5 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$, $0<\lambda \le min\{p,q\}$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{p-\lambda -1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{q-\lambda -1}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{q}{3\lambda n^{\frac{\lambda }{q}}}]n^{p-\lambda -1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{p}{3\lambda n^{\frac{\lambda }{p}}}]n^{q-\lambda -1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.17)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{(p-1)\lambda -1}}{{[k_{\lambda }-\frac{p}{3\lambda n^{\frac{\lambda }{p}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[k_{\lambda }-\frac{q}{3\lambda n^{\frac{\lambda }{q}}}]n^{p-\lambda -1}{a}_n^p,$$

(3.18)

where $k_{\lambda }=\frac{pq}{\lambda }>0$. Inequality (3.17) is equivalent to (3.18). In particular, we have the equivalent inequalities as follows.

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m^{\lambda },n^{\lambda }\}}<k_{\lambda }{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{q}{k_{\lambda }3\lambda n^{\frac{\lambda }{q}}}]n^{p-\lambda -1}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{q-\lambda -1}{b}_n^q\right\}}^{\frac{1}{q}},$$

(3.19)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{(p-1)\lambda -1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m^{\lambda },n^{\lambda }\}}\right]}^p<k_{\lambda }^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{q}{3k_{\lambda }\lambda n^{\frac{\lambda }{q}}}]n^{p-\lambda -1}{a}_n^p.$$

(3.20)

Set $\lambda =1$, combining (3.1) and (3.2), we have the following.

Corollary 3.6 Assume that $p,q>1$, $\frac{1}{p}+\frac{1}{q}=1$,$r>1$, $\frac{1}{r}+\frac{1}{s}=1$, $a_n\ge 0$, $b_n\ge 0$, such that $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{s}-1}{a}_n^p<\mathrm{\infty }$ and $0<{\sum }_{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{b}_n^q<\mathrm{\infty }$, then

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m,n\}}<& {\left\{\sum _{n=1}^{\mathrm{\infty }}[rs-\frac{s}{3n^{\frac{1}{s}}}]n^{\frac{p}{s}-1}{a}_n^p\right\}}^{\frac{1}{p}}\\ \times {\left\{\sum _{n=1}^{\mathrm{\infty }}[rs-\frac{r}{3n^{\frac{1}{r}}}]n^{\frac{q}{r}-1}{b}_n^q\right\}}^{\frac{1}{q}},\end{array}$$

(3.21)

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{\frac{p}{s}-1}}{{[rs-\frac{r}{3n^{\frac{1}{r}}}]}^{p-1}}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m,n\}}\right]}^p<\sum _{n=1}^{\mathrm{\infty }}[rs-\frac{s}{3n^{\frac{1}{s}}}]n^{\frac{p}{s}-1}{a}_n^p.$$

(3.22)

In particular, we have the equivalent inequalities as follows.

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m,n\}}<rs{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{1}{3r{n}^{\frac{1}{s}}}]n^{\frac{p}{s}-1}{a}_n^p\right\}}^{\frac{1}{p}}{\left\{\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{q}{r}-1}{b}_n^q\right\}}^{\frac{1}{q}},$$

(3.23)

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{\frac{p}{s}-1}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m,n\}}\right]}^p<{(rs)}^p\sum _{n=1}^{\mathrm{\infty }}[1-\frac{1}{3r{n}^{\frac{1}{s}}}]n^{\frac{p}{s}-1}{a}_n^p.$$

(3.24)

Taking $p=q=r=s=2$, in (3.23) and (3.24), we have

$$\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{a_m{b}_n}{max\{m,n\}}<4{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{1}{6\sqrt{n}}]a_n^2\right\}}^{\frac{1}{2}}{\left\{\sum _{n=1}^{\mathrm{\infty }}[1-\frac{1}{6\sqrt{n}}]b_n^2\right\}}^{\frac{1}{2}},$$

(3.25)

$$\sum _{n=1}^{\mathrm{\infty }}{\left[\sum _{m=1}^{\mathrm{\infty }}\frac{a_m}{max\{m,n\}}\right]}^2<16\sum _{n=1}^{\mathrm{\infty }}[1-\frac{1}{6\sqrt{n}}]a_n^2.$$

(3.26)

Remark 3.1 For $r=\frac{\lambda p}{\lambda +p-2}$ and $s=\frac{\lambda q}{\lambda +q-2}$ in Theorem 3.1, we get the results of [4].